Question

The first hill of a roller coaster is 42.0 meters high. The roller coaster drops to a height of
4.2 meters above the ground at the bottom of this first hill. If the mass of the roller
coaster is 34,500 kg, what is the velocity at the bottom of the first hill? Ignore friction and
assume the velocity at the top of the first hill is 0 m/s.

Answer 1

Velocity = 27.2 m/s

Explanation:

The velocity at the bottom of the first hill can be determined with the change in Kinetic energy. Due to energy conversion, both the change in kinetic energy and potential energy is same when the height is changed.

So, by applying the formulas:

KE = $\frac{m v^{2}}{2}$

where KE is Kinetic energy, m is the mass and v is the Velocity.

PE = mgh

where PE is Potential energy, m is the mass, g is the gravitational force and h is the height.

Since, both the energies are equal, we can write:

PE = KE

$\frac{m v^{2}}{2}$ = mgh

v²   = 2gh

v     = $\sqrt{2 g h}$

To Substitute:

h = 42 - 4.2 = 37.8 m

g = 9.8 m/s²

So, v = $\sqrt{2 \times 9.8 \times 37.8}$

        = 27.2 m/s